Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{-9x^2 + 54x}{x^2 - 7x + 12} \times \dfrac{5x - 15}{9x - 54} $
Explanation: First factor the quadratic. $n = \dfrac{-9x^2 + 54x}{(x - 3)(x - 4)} \times \dfrac{5x - 15}{9x - 54} $ Then factor out any other terms. $n = \dfrac{-9x(x - 6)}{(x - 3)(x - 4)} \times \dfrac{5(x - 3)}{9(x - 6)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -9x(x - 6) \times 5(x - 3) } { (x - 3)(x - 4) \times 9(x - 6) } $ $n = \dfrac{ -45x(x - 6)(x - 3)}{ 9(x - 3)(x - 4)(x - 6)} $ Notice that $(x - 6)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -45x(x - 6)\cancel{(x - 3)}}{ 9\cancel{(x - 3)}(x - 4)(x - 6)} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $n = \dfrac{ -45x\cancel{(x - 6)}\cancel{(x - 3)}}{ 9\cancel{(x - 3)}(x - 4)\cancel{(x - 6)}} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $n = \dfrac{-45x}{9(x - 4)} $ $n = \dfrac{-5x}{x - 4} ; \space x \neq 3 ; \space x \neq 6 $